Metals and Their Compounds

78 soru

Soru 1Soru

In the industrial extraction of sodium metal using the Downs cell, calcium chloride (CaCl2\text{CaCl}_2) is added to molten sodium chloride (NaCl\text{NaCl}) primarily to:

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Cevap: lower the melting point of the sodium chloride electrolyte

Cevap

Calcium chloride is added to lower the melting point of the sodium chloride electrolyte.
Pure sodium chloride melts at 801C801^\circ\text{C}. Adding calcium chloride lowers the melting point of the mixture to about 600C600^\circ\text{C}, which conserves energy and minimizes evaporation of the liberated sodium metal during electrolysis in the Downs cell.

Adım Adım Çözüm

1
Identify the physical properties of pure sodium chloride
Pure NaCl\text{NaCl} has a high melting point of approximately 801C801^\circ\text{C}.
Maintaining such a high temperature industrially requires significant thermal energy.
2
Determine the effect of adding calcium chloride (CaCl2\text{CaCl}_2)
Mixing CaCl2\text{CaCl}_2 with NaCl\text{NaCl} forms an electrolytic mixture with a reduced melting point of about 600C600^\circ\text{C}.
Lowering the melting point makes the process economically viable and reduces heat loss and electrode wear.

Anahtar Kavram

Function of flux/admixture in the Downs process for sodium extraction
Tahmini Süre:45s
Soru 2Soru

When a few drops of aqueous sodium hydroxide are added to a solution of aluminium chloride, a white gelatinous precipitate forms. Upon adding excess sodium hydroxide, the precipitate dissolves to produce a clear, colorless solution. Which chemical species is present in the final clear solution?

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Cevap: Tetrahydroxoaluminate(III) ion, [Al(OH)4][Al(OH)_4]^-

Cevap

The tetrahydroxoaluminate(III) ion, [Al(OH)4][Al(OH)_4]^-
Aluminium hydroxide, Al(OH)3Al(OH)_3, is an amphoteric hydroxide. It initially precipitates as a white gelatinous solid when alkali is added. When excess strong alkali like sodium hydroxide is added, it acts as an acid and reacts further with hydroxide ions to form the soluble complex ion, tetrahydroxoaluminate(III), [Al(OH)4][Al(OH)_4]^-, yielding a clear solution.

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1
Identify the initial precipitation reaction
Adding a few drops of OHOH^- ions precipitates white gelatinous aluminium hydroxide: Al3+(aq)+3OH(aq)Al(OH)3(s)Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)
Aluminium ions react with hydroxide ions to form insoluble aluminium hydroxide.
2
Apply the amphoteric property of aluminium hydroxide with excess strong base
Adding excess sodium hydroxide causes Al(OH)3Al(OH)_3 to act as an acid, dissolving according to: Al(OH)3(s)+OH(aq)[Al(OH)4](aq)Al(OH)_3(s) + OH^-(aq) \rightarrow [Al(OH)_4]^-(aq)
Amphoteric hydroxides react with excess alkalis to form soluble complex aluminate ions.

Anahtar Kavram

Amphoteric nature of aluminium compounds
Tahmini Süre:1m 0s
Soru 3Soru

Match each aluminium alloy or extraction reagent on the left with its correct composition, primary industrial application, or function on the right.

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Öğeler

Duralumin
Magnalium
Alnico
Cryolite (Na3AlF6Na_3AlF_6)

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Cevap

Duralumin matches with the composition of AlAl, CuCu, MgMg, MnMn used in aircraft bodies; Magnalium matches with the composition of AlAl, MgMg used in balance beams; Alnico matches with the composition of AlAl, NiNi, CoCo, FeFe used in permanent magnets; Cryolite matches with the molten solvent that lowers the melting point of alumina in extraction.
Each item correctly matches its specific chemical composition and technological usage in metallurgy.

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1
Identify the composition and application of Duralumin
Duralumin contains aluminium, copper, magnesium, and manganese, providing high strength and lightness for aircraft construction.
Alloying aluminium with copper and manganese enhances structural strength.
2
Identify the composition and application of Magnalium
Magnalium is an aluminium-magnesium alloy prized for low density and corrosion resistance in optical/scientific instruments.
Magnesium lowers density and improves machinability.
3
Identify the composition and application of Alnico
Alnico is composed of aluminium, nickel, cobalt, and iron, essential for permanent magnets.
The combination of ferromagnetic metals with aluminium yields high magnetic coercivity.
4
Identify the role of Cryolite in electrolysis
Cryolite acts as a molten solvent for alumina to reduce operating energy costs and enhance ionic conductivity.
Pure alumina melts at over 2000C2000^\circ C; cryolite lowers this operating temperature to around 950C950^\circ C.

Anahtar Kavram

Aluminium Alloys and Extraction Metallurgy
Soru 4Soru

During the industrial extraction of iron from hematite (Fe2O3\text{Fe}_2\text{O}_3) in the blast furnace, which chemical species acts as the primary reducing agent responsible for converting the iron ore to iron in the upper region of the furnace?

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Cevap: Carbon(II) oxide (CO\text{CO})

Cevap

Carbon(II) oxide (CO\text{CO})
In the blast furnace, carbon(II) oxide (CO\text{CO}) gas is produced when carbon dioxide reacts with red-hot coke. Because it is a gas, carbon(II) oxide thoroughly mixes with and reduces the solid hematite (Fe2O3\text{Fe}_2\text{O}_3) to iron in the upper, cooler region of the furnace.

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1
Identify the chemical reactions occurring in the blast furnace.
Coke burns in oxygen to form CO2\text{CO}_2, which then reacts with excess hot coke to produce CO\text{CO}: CO2(g)+C(s)2CO(g)\text{CO}_2(g) + \text{C}(s) \rightarrow 2\text{CO}(g).
Gaseous CO\text{CO} is generated to serve as the gaseous reducing agent.
2
Determine which species reduces hematite in the upper zone (400°C – 700°C).
Gaseous carbon(II) oxide reduces Fe2O3\text{Fe}_2\text{O}_3: Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g).
Gaseous CO\text{CO} provides intimate surface contact with the solid ore compared to solid coke.

Anahtar Kavram

Blast furnace reduction of iron ore by carbon(II) oxide gas
Soru 5Soru

During the industrial extraction of iron in a blast furnace, hematite (Fe2O3Fe_2O_3) is reduced to iron in the upper region of the furnace. Which chemical species acts as the primary reducing agent in this upper zone?

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Cevap: Carbon(II) oxide (COCO)

Cevap

Carbon(II) oxide (COCO)
In the blast furnace, carbon(II) oxide (COCO) gas is produced when hot carbon dioxide reacts with excess coke. Rising COCO gas encounters descending hematite (Fe2O3Fe_2O_3) in the upper, cooler zone of the furnace and reduces it to iron: Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g).

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1
Identify the chemical reactions occurring in the upper region of the blast furnace.
Near the top of the furnace (200°C–700°C), carbon(II) oxide gas reacts with descending hematite ore.
Gas-solid contact allows COCO to easily reduce the porous iron ore.
2
Write the overall reduction equation.
Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)
This chemical equation confirms that COCO gains oxygen (is oxidized) while reducing Fe2O3Fe_2O_3 to metallic iron.

Anahtar Kavram

Role of gaseous carbon(II) oxide as the principal reducing agent in blast furnace iron extraction
Tahmini Süre:45s
Soru 6Soru

Match each chemical substance or process associated with iron extraction and rust prevention on the left with its corresponding chemical role or function on the right.

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Öğeler

Limestone (CaCO3\text{CaCO}_3)
Carbon monoxide (CO\text{CO})
Galvanization
Calcium silicate (CaSiO3\text{CaSiO}_3)

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Cevap

Limestone (CaCO₃) matches with thermal decomposition to provide calcium oxide as a basic flux. Carbon monoxide (CO) matches with serving as the main reducing agent. Galvanization matches with sacrificial coating of iron using zinc. Calcium silicate (CaSiO₃) matches with forming molten slag that floats on molten iron.
Limestone decomposes into calcium oxide, which acts as a basic flux to neutralize silica impurities. Carbon monoxide is the main gaseous reducing agent reducing hematite to metallic iron. Galvanization applies a protective, sacrificial layer of zinc onto iron surfaces. Calcium silicate forms the molten slag layer that sits above molten iron to protect it from re-oxidation.

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1
Identify the role of limestone in the blast furnace.
Limestone undergoes endothermic decomposition to form calcium oxide (CaO\text{CaO}), acting as a basic flux.
Flux is required to react with acidic impurities like silicon(IV) oxide.
2
Identify the primary reducing agent in iron extraction.
Carbon monoxide (CO\text{CO}) reduces hematite (Fe2O3\text{Fe}_2\text{O}_3) to iron in the upper and middle zones of the furnace.
At elevated blast furnace temperatures, gaseous carbon monoxide readily abstracts oxygen from iron ores.
3
Determine the role of zinc coating on iron (galvanization).
Galvanization provides sacrificial protection against corrosion.
Zinc oxidizes preferentially to iron because of its higher position in the electrochemical series.
4
Determine the identity and function of slag.
Calcium silicate (CaSiO3\text{CaSiO}_3) constitutes molten slag.
Slag is less dense than liquid iron, floating on top to prevent oxidation by incoming air blasts.

Anahtar Kavram

Industrial extraction of iron in the blast furnace and sacrificial protection mechanisms against iron rusting.
Soru 7Soru

Match each transition metal or transition metal compound listed on the left with its corresponding industrial catalytic process on the right.

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Öğeler

Finely divided iron (FeFe)
Vanadium(V) oxide (V2O5V_2O_5)
Nickel (NiNi)
Platinum (PtPt)

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Cevap

Finely divided iron matches the Haber process for manufacturing ammonia; Vanadium(V) oxide matches the Contact process for manufacturing tetraoxosulfate(VI) acid; Nickel matches the hydrogenation of vegetable oils to margarine; Platinum matches the Ostwald process for manufacturing trioxonitrate(V) acid.
Transition metals and their oxides serve as effective industrial catalysts due to their partially filled d-orbitals, variable oxidation states, and ability to adsorb reactant molecules onto their surfaces. Finely divided iron is the standard catalyst in the Haber process for ammonia synthesis, vanadium(V) oxide catalyzes sulfur dioxide oxidation in the Contact process, nickel catalyzes the hydrogenation of unsaturated vegetable oils, and platinum catalyzes the catalytic oxidation of ammonia in the Ostwald process.

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1
Identify the industrial reaction associated with finely divided iron.
Finely divided iron catalyzes N2(g)+3H2(g)2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) in the Haber process.
Iron provides a surface for nitrogen and hydrogen molecules to adsorb and react efficiently.
2
Identify the catalyst used in the Contact process.
Vanadium(V) oxide (V2O5V_2O_5) catalyzes 2SO2(g)+O2(g)2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g).
Vanadium exhibits variable oxidation states (V5+V^{5+} and V4+V^{4+}) allowing intermediate redox steps.
3
Identify the catalyst used in organic hydrogenation.
Nickel (NiNi) catalyzes the conversion of unsaturated vegetable oils to saturated fats.
Finely divided nickel adsorbs hydrogen gas and liquid oil to facilitate addition across carbon-carbon double bonds.
4
Identify the catalyst used in the Ostwald process.
Platinum (PtPt) catalyzes the oxidation of ammonia (NH3NH_3) to nitrogen(II) oxide (NONO).
Platinum gauze provides high surface area and stability at elevated temperatures required for ammonia oxidation.

Anahtar Kavram

Industrial Catalytic Applications of Transition Metals
Soru 8Soru

Copper is a transition metal with an atomic number of 2929. What is the ground-state electronic configuration of the copper(II) ion (Cu2+Cu^{2+}) present in copper compounds such as copper(II) tetraoxosulfate(VI)?

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Cevap: [Ar]3d9[Ar] 3d^9

Cevap

The ground-state electronic configuration of the copper(II) ion (Cu2+Cu^{2+}) is [Ar]3d9[Ar] 3d^9.
The correct answer specifies [Ar]3d9[Ar] 3d^9. Neutral copper has a ground-state configuration of [Ar]3d104s1[Ar] 3d^{10} 4s^1. Ionization to form Cu2+Cu^{2+} requires losing two electrons; the first comes from the outermost 4s4s orbital, and the second comes from the 3d3d subshell, leaving [Ar]3d9[Ar] 3d^9.

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1
Determine the electronic configuration of a neutral copper atom (Cu,Z=29Cu, Z=29).
Neutral copper has the ground-state configuration [Ar]3d104s1[Ar] 3d^{10} 4s^1 (anomalous filling for extra stability of a full d-subshell).
A completely filled 3d103d^{10} subshell provides lower overall energy than a 3d94s23d^9 4s^2 arrangement.
2
Identify which electrons are removed when forming the Cu2+Cu^{2+} ion.
Two electrons must be removed: 1 electron from the 4s4s orbital and 1 electron from the 3d3d orbital.
Electrons in the outermost principal quantum level (n=4n=4) are always removed first during cation formation.
3
Write the final electronic configuration of Cu2+Cu^{2+}.
[Ar]3d9[Ar] 3d^9
Subtracting 11 electron from 4s4s and 11 electron from 3d103d^{10} leaves 99 electrons in the 3d3d subshell and 00 in 4s4s.

Anahtar Kavram

Electronic configuration of transition metal cations and anomalous electron filling in copper.
Tahmini Süre:1m 0s
Soru 9Soru

Match each metallic alloy listed in Column A with its corresponding elemental composition and primary application in Column B.

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Öğeler

Duralumin
Brass
Stainless Steel
Bronze

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Cevap

Duralumin matches with Aluminum, Copper, Magnesium, and Manganese (aircraft construction); Brass matches with Copper and Zinc (musical instruments and fittings); Stainless Steel matches with Iron, Carbon, Chromium, and Nickel (cutlery and surgical tools); Bronze matches with Copper and Tin (statues, bearings, and medals).
Each alloy is paired strictly according to its constituent metals and major application: Duralumin (Al+Cu+Mg+Mn\text{Al}+\text{Cu}+\text{Mg}+\text{Mn}) for light aircraft structures, Brass (Cu+Zn\text{Cu}+\text{Zn}) for acoustic instruments, Stainless Steel (Fe+C+Cr+Ni\text{Fe}+\text{C}+\text{Cr}+\text{Ni}) for rust-proof tools, and Bronze (Cu+Sn\text{Cu}+\text{Sn}) for low-friction bearings and statues.

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1
Analyze the composition and primary characteristic of Duralumin.
Duralumin consists of Al+Cu+Mg+Mn\text{Al} + \text{Cu} + \text{Mg} + \text{Mn}. Its key physical property is low density combined with high tensile strength.
Aluminum is the base metal in lightweight aviation alloys.
2
Differentiate between the copper-based alloys: Brass and Bronze.
Brass is an alloy of copper and zinc (Cu+Zn\text{Cu} + \text{Zn}), whereas Bronze is an alloy of copper and tin (Cu+Sn\text{Cu} + \text{Sn}).
Zinc is added to copper to make brass; tin is added to copper to make bronze.
3
Identify the alloying elements that impart corrosion resistance to steel.
Stainless steel contains chromium and nickel added to iron and carbon.
Chromium reacts with oxygen to form a thin, unreactive passivation layer of chromium(III) oxide (Cr2O3\text{Cr}_2\text{O}_3).

Anahtar Kavram

Compositions, physical property modifications, and industrial applications of major metallic alloys.
Soru 10Soru

During the industrial extraction of iron in the blast furnace, limestone (CaCO3\text{CaCO}_3) is decomposed by heat to form calcium oxide (CaO\text{CaO}), which then reacts with silica (SiO2\text{SiO}_2) impurities present in the ore. Which of the following chemical formulas represents the molten slag produced from this reaction?

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Cevap: CaSiO3\text{CaSiO}_3

Cevap

CaSiO3\text{CaSiO}_3 (Calcium trioxosilicate(IV))
In the lower region of the blast furnace, limestone (CaCO3\text{CaCO}_3) decomposes into calcium oxide (CaO\text{CaO}). The basic CaO\text{CaO} reacts with acidic silica (SiO2\text{SiO}_2) impurities present in hematite to form molten calcium trioxosilicate(IV) (CaSiO3\text{CaSiO}_3), commonly known as slag.

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1
Identify thermal decomposition of limestone
CaCO3(s)CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)
High temperatures in the blast furnace break down limestone into basic calcium oxide.
2
Combine basic oxide flux with acidic silica impurity
CaO(s)+SiO2(s)CaSiO3(l)\text{CaO}(s) + \text{SiO}_2(s) \rightarrow \text{CaSiO}_3(l)
Calcium oxide acts as a basic flux that neutralizes acidic sand/silica impurities to form molten slag.

Anahtar Kavram

Slag Formation in Iron Extraction
Tahmini Süre:45s
Soru 11Soru

During the electrolytic refining of copper, a steady current of 5.0 A5.0\text{ A} is passed through an aqueous copper(II) tetraoxosulfate(VI) (CuSO4CuSO_4) solution for 965 seconds965\text{ seconds}. What mass of copper, in grams, is deposited at the cathode? (Faraday's constant F=96,500 C mol1F = 96,500\text{ C mol}^{-1}; Molar mass of Cu=64 g mol1Cu = 64\text{ g mol}^{-1})

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Cevap: 1.6

Cevap

1.6 g
Passing a 5.0 A5.0\text{ A} current for 965 s965\text{ s} transfers 4825 C4825\text{ C} of electric charge, corresponding to 0.05 mol0.05\text{ mol} of electrons. Because the deposition of copper from CuSO4CuSO_4 follows Cu2++2eCuCu^{2+} + 2e^- \rightarrow Cu, every 2 moles2\text{ moles} of electrons deposit 1 mole1\text{ mole} of copper metal. Thus, 0.025 mol0.025\text{ mol} of copper is deposited, which corresponds to 0.025 mol×64 g mol1=1.6 g0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}.

Adım Adım Çözüm

1
Calculate the total quantity of electricity (QQ) passed through the electrolyte.
Q=I×t=5.0 A×965 s=4825 CQ = I \times t = 5.0\text{ A} \times 965\text{ s} = 4825\text{ C}
Electric charge is the product of current in amperes and time in seconds.
2
Calculate the moles of electrons transferred.
n(e)=QF=4825 C96500 C mol1=0.05 moln(e^-) = \frac{Q}{F} = \frac{4825\text{ C}}{96500\text{ C mol}^{-1}} = 0.05\text{ mol} of electrons
One mole of electrons carries a charge equivalent to 1 Faraday (96,500 C96,500\text{ C}).
3
Use the cathode half-equation to find the moles of deposited copper.
Cathode reaction: Cu(aq)2++2eCu(s)Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}. Moles of Cu=0.05 mol2=0.025 molCu = \frac{0.05\text{ mol}}{2} = 0.025\text{ mol}
Reduction of one mole of copper(II) ions requires two moles of electrons.
4
Convert the moles of deposited copper into mass.
Mass of Cu=n×M=0.025 mol×64 g mol1=1.6 gCu = n \times M = 0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}
Multiplying the chemical amount of copper by its molar mass yields the mass in grams.

Anahtar Kavram

Quantitative electrolysis of copper(II) ions using Faraday's laws of electrolysis
Soru 12Soru

What is the ground-state electronic configuration of the iron(III) ion, Fe3+Fe^{3+}? (Atomic number of Fe=26Fe = 26)

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Cevap: [Ar]3d5[Ar] 3d^5

Cevap

The correct ground-state electronic configuration of the iron(III) ion is [Ar]3d5[Ar] 3d^5.
Neutral iron (Z=26Z=26) has the ground-state electron configuration [Ar]3d64s2[Ar] 3d^6 4s^2. When transition metal atoms form cations, electrons are removed first from the outermost principal energy level (4s4s) before removing electrons from the inner 3d3d subshell. Ionization to Fe3+Fe^{3+} involves removing the two 4s4s electrons and one 3d3d electron, resulting in a stable half-filled dd-subshell configuration of [Ar]3d5[Ar] 3d^5.

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1
Determine the electronic configuration of the neutral iron atom (FeFe, Z=26Z = 26).
Neutral FeFe has 26 electrons: 1s22s22p63s23p63d64s21s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2 or abbreviated as [Ar]3d64s2[Ar] 3d^6 4s^2.
The 4s4s orbital is filled before 3d3d in neutral atoms according to the Aufbau principle.
2
Apply the cation ionization rule for transition metals to form Fe3+Fe^{3+}.
Remove 3 electrons in total: first remove 2 electrons from the outermost 4s4s orbital, then remove 1 electron from the 3d3d orbital.
Electrons in the outermost shell (n=4n=4) are lost before inner (n1)d(n-1)d electrons during ionization.
3
Write the resulting electronic configuration for Fe3+Fe^{3+}.
[Ar]3d5[Ar] 3d^5
Removing two 4s4s electrons leaves [Ar]3d6[Ar] 3d^6, and removing one more 3d3d electron yields [Ar]3d5[Ar] 3d^5.

Anahtar Kavram

Electronic configuration of transition metal cations
Soru 13Soru

Match each aluminium extraction component, compound, or alloy listed on the left with its accurate chemical role, structural behavior, or application on the right.

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Öğeler

Molten Cryolite (Na3AlF6\text{Na}_3\text{AlF}_6) in the Hall-Héroult cell
Reaction of bauxite (Al2O32H2O\text{Al}_2\text{O}_3\cdot 2\text{H}_2\text{O}) with concentrated NaOH(aq)\text{NaOH}_{(aq)}
Anhydrous aluminium chloride (AlCl3\text{AlCl}_3) below 400C400^\circ\text{C}
Duralumin composition and application

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Cevap

Molten Cryolite matches with its role as a solvent lowering the melting point of alumina and improving conductivity; reaction of bauxite with concentrated alkali matches with the formation of soluble sodium tetrahydroxoaluminate(III); anhydrous aluminium chloride below 400 °C matches with its existence as a covalent dimer Al2Cl6 with dative bonds; Duralumin matches with the alloy composed of Al, Cu, Mn, Mg used in aircraft structure.
Cryolite serves as a flux and solvent reducing alumina's melting point from 2050C2050^\circ\text{C} to 950C950^\circ\text{C}. The reaction of bauxite with concentrated sodium hydroxide exploits aluminium's amphoteric property to produce soluble tetrahydroxoaluminate(III). Anhydrous aluminium chloride forms a coordinate-bonded dimer (Al2Cl6\text{Al}_2\text{Cl}_6) at low temperatures. Duralumin is an aluminium-copper-manganese-magnesium alloy key to aerospace applications due to its high strength-to-weight ratio.

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1
Analyze the role of cryolite in industrial extraction
Cryolite lowers the melting temperature of Al2O3\text{Al}_2\text{O}_3 from 2050C2050^\circ\text{C} to 950C950^\circ\text{C} and increases conductivity.
Pure Al2O3\text{Al}_2\text{O}_3 has an extremely high melting point and poor conductivity in the solid state; cryolite provides a suitable molten electrolyte mixture.
2
Examine the Bayer process chemical separation of amphoteric aluminium oxide
Dissolution in concentrated NaOH\text{NaOH} forms soluble complex ion Na[Al(OH)4]\text{Na}[\text{Al}(\text{OH})_4].
Aluminium oxide is amphoteric and reacts with strong base, whereas basic impurities like Fe2O3\text{Fe}_2\text{O}_3 remain insoluble.
3
Evaluate the molecular structure of anhydrous aluminium chloride
Forms Al2Cl6\text{Al}_2\text{Cl}_6 dimer via coordinate bonds.
Aluminium in monomeric AlCl3\text{AlCl}_3 has only 6 valence electrons; dimerization allows each aluminium atom to complete its octet.
4
Identify the composition and application of Duralumin
Alloy of Al\text{Al}, Cu\text{Cu}, Mn\text{Mn}, Mg\text{Mg} used in aircraft.
Addition of copper and magnesium to aluminium dramatically increases mechanical strength without significantly compromising low density.

Anahtar Kavram

Chemical principles of aluminium extraction (Bayer and Hall-Héroult processes), amphoteric reactivity, dimerization of aluminium chloride, and metallurgical properties of aluminium alloys.
Soru 14Soru

Match each transition metal complex species on the left with its corresponding IUPAC name and oxidation state of the central metal ion on the right.

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Öğeler

[Fe(CN)6]3[Fe(CN)_6]^{3-}
[Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}
[Co(H2O)6]2+[Co(H_2O)_6]^{2+}
[Ni(CO)4][Ni(CO)_4]

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Cevap

The correct matches are: [Fe(CN)6]3[Fe(CN)_6]^{3-} matches Hexacyanoferrate(III) ion (FeFe in +3+3 state); [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+} matches Tetraamminecopper(II) ion (CuCu in +2+2 state); [Co(H2O)6]2+[Co(H_2O)_6]^{2+} matches Hexaaquacobalt(II) ion (CoCo in +2+2 state); and [Ni(CO)4][Ni(CO)_4] matches Tetracarbonylnickel(0) (NiNi in 00 state).
Each complex is correctly paired based on the number and charge of its ligands, the resulting oxidation state of the central transition metal, and IUPAC nomenclature rules for cationic, neutral, and anionic complexes.

Adım Adım Çözüm

1
Determine the charge of each ligand present in the transition metal complex.
Cyano (CNCN^-) carries a charge of 1-1, while ammine (NH3NH_3), aqua (H2OH_2O), and carbonyl (COCO) are neutral (00 charge).
Knowing ligand charges is required to set up the algebraic equation for the metal's oxidation state.
2
Calculate the oxidation state of the central metal ion in each complex.
For [Fe(CN)6]3[Fe(CN)_6]^{3-}: x+6(1)=3x=+3x + 6(-1) = -3 \Rightarrow x = +3. For [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}: x+4(0)=+2x=+2x + 4(0) = +2 \Rightarrow x = +2. For [Co(H2O)6]2+[Co(H_2O)_6]^{2+}: x+6(0)=+2x=+2x + 6(0) = +2 \Rightarrow x = +2. For [Ni(CO)4][Ni(CO)_4]: x+4(0)=0x=0x + 4(0) = 0 \Rightarrow x = 0.
The sum of oxidation numbers of the central atom and ligands equals the total net charge of the complex species.
3
Apply standard IUPAC naming rules for complex species.
Name ligands with multiplicative prefixes (tetra-, hexa-), follow with the metal name (using '-ate' suffix for anionic complexes like ferrate), and indicate the oxidation state in Roman numerals.
Anionic complexes modify the metal name root, whereas cationic and neutral complexes retain the standard metal element name.

Anahtar Kavram

IUPAC Nomenclature and Oxidation State Calculations for Transition Metal Complex Ions
Soru 15Soru

When copper metal turnings are heated with concentrated tetraoxosulfate(VI) acid (H2SO4H_2SO_4), a dense gas is evolved alongside the formation of copper(II) tetraoxosulfate(VI) (CuSO4CuSO_4) and water (H2OH_2O). Which of the following correctly identifies the gaseous product evolved and the specific role of concentrated H2SO4H_2SO_4 in this chemical reaction?

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Cevap: Sulfur(IV) oxide (SO2SO_2); concentrated H2SO4H_2SO_4 acts as an oxidizing agent.

Cevap

Sulfur(IV) oxide (SO2SO_2); concentrated H2SO4H_2SO_4 acts as an oxidizing agent.
Copper is an unreactive transition metal located below hydrogen in the electrochemical series. Consequently, it does not react with dilute acids to liberate hydrogen gas. However, when heated with concentrated tetraoxosulfate(VI) acid, the acid behaves as a strong oxidizing agent rather than a typical acid. Copper is oxidized to copper(II) tetraoxosulfate(VI) (CuSO4CuSO_4), while the sulfur in H2SO4H_2SO_4 is reduced from oxidation state +6+6 to +4+4, liberating sulfur(IV) oxide gas (SO2SO_2) along with water.

Adım Adım Çözüm

1
Analyze the position of copper in the electrochemical activity series.
Copper is less electropositive than hydrogen (ECu2+/Cu=+0.34 VE^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}), meaning it cannot displace H2H_2 gas from dilute or non-oxidizing acids.
Direct single replacement of hydrogen by copper is thermodynamically unfavorable.
2
Determine the chemical action of hot concentrated tetraoxosulfate(VI) acid on copper.
Hot concentrated H2SO4H_2SO_4 is a strong oxidizing agent. It oxidizes copper from oxidation state 00 to +2+2 (Cu2+Cu^{2+}).
Oxidizing acids react with unreactive metals via redox pathways rather than acid-base displacement.
3
Formulate the balanced thermochemical equation.
Cu(s)+2H2SO4(aq)CuSO4(aq)+2H2O(l)+SO2(g)Cu_{(s)} + 2H_2SO_{4(aq)} \rightarrow CuSO_{4(aq)} + 2H_2O_{(l)} + SO_{2(g)}
Sulfur in H2SO4H_2SO_4 (oxidation state +6+6) undergoes reduction to form sulfur(IV) oxide (SO2SO_2, oxidation state +4+4).

Anahtar Kavram

Chemical properties of copper and oxidizing behavior of concentrated tetraoxosulfate(VI) acid
Tahmini Süre:2m 0s
Soru 16Soru

Arrange the following sequential industrial steps of the Bayer process used in refining bauxite ore into pure alumina (aluminium oxide) in the correct chronological order from first to last.

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The correct chronological order of the Bayer process is: Digestion of bauxite in concentrated NaOH \rightarrow Filtration to remove red mud \rightarrow Seeding to precipitate hydrated aluminium hydroxide \rightarrow High-temperature calcination to yield pure anhydrous alumina.
The Bayer process refines crude bauxite into pure alumina through four distinct chemical phases: initial amphoteric dissolution in hot concentrated alkali (digestion), removal of solid iron oxide impurities via filtration (red mud removal), controlled crystallization of pure hydroxide (seeding/precipitation), and thermal decomposition (calcination) to produce anhydrous Al2O3\text{Al}_2\text{O}_3.

Adım Adım Çözüm

1
Identify the chemical extraction reaction
Crude bauxite (Al2O3xH2O\text{Al}_2\text{O}_3\cdot x\text{H}_2\text{O}) is treated with hot concentrated NaOH\text{NaOH} solution under pressure to dissolve aluminium amphoterically: Al2O3(s)+2NaOH(aq)+3H2O(l)2NaAl(OH)4(aq)\text{Al}_2\text{O}_3(s) + 2\text{NaOH}(aq) + 3\text{H}_2\text{O}(l) \rightarrow 2\text{NaAl(OH)}_4(aq).
Aluminium oxide is amphoteric and forms soluble aluminate ions, whereas impurities like Fe2O3\text{Fe}_2\text{O}_3 do not react.
2
Separate insoluble residues
The dense, insoluble residue known as 'red mud' (containing Fe2O3\text{Fe}_2\text{O}_3, silica, and titania) is filtered out.
Filtration purifies the liquid stream so that subsequent precipitates are free from iron contamination.
3
Precipitate hydrated aluminium hydroxide
The clear sodium tetrahydroxoaluminate filtrate is cooled and seeded with pure Al(OH)3\text{Al(OH)}_3 crystals to induce precipitation: NaAl(OH)4(aq)Al(OH)3(s)+NaOH(aq)\text{NaAl(OH)}_4(aq) \rightarrow \text{Al(OH)}_3(s) + \text{NaOH}(aq).
Cooling and seeding shifts the equilibrium back toward solid Al(OH)3\text{Al(OH)}_3 formation.
4
Dehydrate the precipitate via heating
The collected Al(OH)3\text{Al(OH)}_3 is washed and calcined in rotary kilns at 1000C1000^\circ\text{C}: 2Al(OH)3(s)ΔAl2O3(s)+3H2O(g)2\text{Al(OH)}_3(s) \xrightarrow{\Delta} \text{Al}_2\text{O}_3(s) + 3\text{H}_2\text{O}(g).
Calcination removes all chemically bound water, producing pure dry alumina feed for the Hall-Héroult electrolytic cell.

Anahtar Kavram

Bayer Process for Bauxite Purification
Soru 17Soru

Soft solder is an alloy commonly used in electrical wiring and plumbing to join metal components. Which pair of metals constitutes soft solder, and what structural mechanism explains why solder is harder than its pure constituent metals?

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Cevap: Lead and tin; foreign atoms of different sizes disrupt the regular metallic lattice, making it harder for layers of metal ions to slide past one another.

Cevap

Soft solder consists of lead and tin. Alloying introduces atoms of a different atomic size into the metallic lattice, creating distortion that restricts the movement of atomic layers, resulting in enhanced hardness.
Soft solder is an alloy of lead and tin. The addition of foreign atoms of differing atomic radii disrupts the regular, symmetrical arrangement of the pure metal lattice. This lattice distortion prevents atomic layers from sliding easily over one another, rendering the alloy harder than pure lead or pure tin.

Adım Adım Çözüm

1
Identify the elemental composition of soft solder.
Soft solder is a binary alloy composed of lead (PbPb) and tin (SnSn).
Memorization and recognition of core industrial alloy compositions specified in the JAMB Chemistry syllabus.
2
Analyze the physical property changes resulting from metallic alloying.
The incorporation of foreign atoms disrupts the uniform layers of the host metal lattice.
Atoms of different sizes introduce lattice strain and distortion, making it significantly more difficult for atomic planes to slide over one another when shear stress is applied.

Anahtar Kavram

Composition of solder and structural explanation of enhanced hardness in alloys due to lattice distortion
Soru 18Soru

Duralumin, an alloy primarily composed of aluminium along with copper, magnesium, and manganese, exhibits higher tensile strength than pure aluminium because the presence of atoms of different sizes distorts the metallic lattice and hinders the slipping of crystal planes.

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Cevap: True

Cevap

The statement is True. Duralumin consists of aluminium alloyed with copper, magnesium, and manganese. The introduction of atoms of different sizes distorts the regular metallic lattice of aluminium, which impedes the movement of atomic planes and increases overall tensile strength.
The statement is true because Duralumin is formed by alloying aluminium with copper, magnesium, and manganese. The differing atomic sizes of these constituent elements cause structural lattice distortion, which prevents atomic layers from sliding easily past each other and enhances tensile strength.

Adım Adım Çözüm

1
Verify the elemental composition of Duralumin.
Duralumin is an aluminium-based alloy containing approximately 94% aluminium, 4% copper, 1% magnesium, and 0.5-1% manganese.
Confirming the constituent metals verifies the composition stated in the stem.
2
Analyze the structural effect of alloying on the metallic crystal lattice.
Copper, magnesium, and manganese atoms have different atomic sizes compared to aluminium atoms, causing localized distortion in the metallic lattice.
Disrupting the uniform arrangement of identical host atoms alters the physical behavior of the metal matrix.
3
Relate lattice distortion to mechanical property modifications (tensile strength).
Lattice distortion creates stress fields that impede the movement of dislocations and slipping of crystal layers under applied force, resulting in higher tensile strength.
Interference with layer sliding directly explains why alloys are stronger and harder than their pure constituent metals.

Anahtar Kavram

Alloy composition and strengthening mechanism via lattice distortion
Soru 19Soru

Match each chemical phenomenon or process involving iron and its compounds listed on the left with its corresponding chemical principle or characteristic observation on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Galvanizing iron structural beams with a thin coating of zinc metal
Addition of aqueous sodium hydroxide (NaOH\text{NaOH}) to iron(II) tetraoxosulfate(VI) solution
Accumulation of molten slag (CaSiO3\text{CaSiO}_3) at the hearth of the blast furnace
Reaction of aqueous iron(II) ions with acidified potassium tetraoxomanganate(VII)

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

1. Galvanizing iron structural beams matches with providing sacrificial cathodic protection due to higher electropositivity of zinc.
2. Addition of aqueous sodium hydroxide to iron(II) tetraoxosulfate(VI) matches with forming a dirty-green precipitate that turns reddish-brown in air.
3. Accumulation of molten slag at the blast furnace hearth matches with floating on molten iron to prevent re-oxidation.
4. Reaction of aqueous iron(II) ions with acidified potassium tetraoxomanganate(VII) matches with decolorizing the purple solution via redox reaction.
Each pair correctly connects an iron chemical phenomenon with its true underlying property: zinc sacrificial protection relies on standard electrode potential differences; Fe2+\text{Fe}^{2+} precipitation produces dirty-green Fe(OH)2\text{Fe(OH)}_2 that oxidizes to brown Fe(OH)3\text{Fe(OH)}_3; slag (CaSiO3\text{CaSiO}_3) protects extracted molten iron from re-oxidation at the furnace base; and Fe2+\text{Fe}^{2+} reduces purple MnO4\text{MnO}_4^- to colorless Mn2+\text{Mn}^{2+}.

Adım Adım Çözüm

1
Analyze the principle of rusting prevention via galvanization
Zinc is more reactive (more electropositive) than iron, so it corrodes preferentially in an electrochemically sacrificial manner.
Protective coatings composed of metals above iron in the electrochemical series function sacrificially.
2
Identify qualitative test reactions for iron(II) ions with strong bases
Adding OH\text{OH}^- ions to Fe2+\text{Fe}^{2+} forms insoluble dirty-green Fe(OH)2\text{Fe(OH)}_2, which oxidizes in air to hydrated iron(III) oxide/hydroxide.
Iron(II) compounds undergo atmospheric oxidation rapidly in alkaline media.
3
Evaluate the industrial function of slag in the blast furnace hearth
Molten CaSiO3\text{CaSiO}_3 forms an immiscible layer above liquid iron due to density differences, preventing oxygen in incoming air blasts from re-oxidizing the extracted metal.
Physical separation of hot molten iron from oxidative gases is crucial to preserve yield.
4
Examine redox properties of iron(II) species with standard oxidizing agents
Fe2+\text{Fe}^{2+} is oxidized to Fe3+\text{Fe}^{3+}, while purple MnO4\text{MnO}_4^- is reduced to colorless Mn2+\text{Mn}^{2+} in acidic solution.
Potassium tetraoxomanganate(VII) is a strong oxidizing agent used to confirm reducing species like Fe2+\text{Fe}^{2+}.

Anahtar Kavram

Chemical reactivity, industrial extractions, qualitative identification, and corrosion mechanisms of iron and its compounds
Soru 20Soru

In the industrial extraction of iron from haematite (Fe2O3Fe_2O_3) in a blast furnace, which chemical species acts as the primary reducing agent for reducing iron(III) oxide in the upper zone of the furnace?

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Cevap: Carbon(II) oxide (COCO)

Cevap

Carbon(II) oxide (COCO)
Gaseous carbon(II) oxide (COCO) is the primary reducing agent in the upper, cooler region of the blast furnace. It reduces iron(III) oxide (Fe2O3Fe_2O_3) to iron according to the reaction equation Fe2O3(s)+3CO(g)2Fe+3CO2(g)Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe + 3CO_2(g).

Adım Adım Çözüm

1
Identify the chemical reactions producing reducing species in the blast furnace.
Coke reacts with oxygen to form carbon(IV) oxide, which is subsequently reduced by excess hot coke higher up in the furnace to yield carbon(II) oxide (CO2+C2COCO_2 + C \rightarrow 2CO).
Gaseous carbon(II) oxide rises through the furnace and comes into intimate contact with descending iron ore.
2
Analyze the reduction step of haematite in the upper zone.
Gaseous COCO reduces Fe2O3Fe_2O_3 via the reaction Fe2O3(s)+3CO(g)2Fe(s/l)+3CO2(g)Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s/l) + 3CO_2(g).
Carbon(II) oxide is a powerful gaseous reducing agent at the moderate temperatures (400°C–700°C) found near the top of the furnace.

Anahtar Kavram

General Principles of Metallurgy and Metal Extraction - Reduction of Iron Ore in the Blast Furnace
Tahmini Süre:1m 0s
Sayfa 1 / 4Sonraki
Metals and Their Compounds Alıştırma Soruları — JAMB UTME | Examkin